Variation in g With Depth
Category : JEE Main & Advanced
Acceleration due to gravity at the surface of the earth
\[g=\frac{GM}{{{R}^{2}}}=\frac{4}{3}\pi \rho GR\] ...(i)
Acceleration due to gravity at depth d from the surface of the earth
\[{g}'=\frac{4}{3}\pi \rho G(R-d)\] ...(ii)
From (i) and (ii) \[{g}'=g\left[ 1-\frac{d}{R} \right]\]
(i) The value of g decreases on going below the surface of the earth. From equation (ii) we get \[{g}'\propto (R-d)\].
So it is clear that if d increase, the value of g decreases.
(ii) At the centre of earth \[\frac{g\,{{R}^{2}}{{T}^{2}}}{4{{\pi }^{2}}}={{\left( R+h \right)}^{3}}\] \[\therefore \ \ {g}'=0\], i.e., the acceleration due to gravity at the centre of earth becomes zero.
(iii) Decrease in the value of g with depth
Absolute decrease \[\Delta g=g-{g}'=\frac{dg}{R}\]
Fractional decrease \[\frac{\Delta g}{g}=\frac{g-{g}'}{g}=\frac{d}{R}\]
Percentage decrease \[\frac{\Delta g}{g}\times 100%=\frac{d}{R}\times 100%\]
(iv) The rate of decrease of gravity outside the earth (\[\text{if}\,\ h<<R\]) is double to that of inside the earth.
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